Maths - Projections of lines on planes

The orientation of the plane is defined by its normal vector B as described
here.

To do this we will use the following notation:

A || B = the component of line A that is projected onto plane B, in other
words a vector to the point on the plane where, if you take a normal at that
point, it will intercept the end of vector A.

A B = the component of line
A is projected onto the normal of plane B.

We will also use:

|A| = the scalar magnitude of vector A.

θ = the angle between vector
A and the normal to plane B

= a unit vector in the direction
of A

So: = A / |A|

Relationship between these quantities

If we add the the parallel and perpendicular components then we get the original
vector, which gives us the following equation:

A = A || B + A B

So if we have the perpendicular component we can work out the parallel component
and visa-versa.

Calculation of the projection on the plane

From the above diagram, the scalar magnitude of the projection on the plane
is |A| sin(θ) and its direction is
along the plane (which is perpendicular to the normal B).

To find the direction that we want, first take a vector which is mutually perpendicular
to A and B, this is given by the cross product A x B (which is out of the page
on the above diagram). Now take a vector which is mutually perpendicular to
this and vector B, this gives us the direction that we want.

So the direction is:

B × (A × B)

or: (B × A) × B

We need to normalise this, so a unit vector in the required direction is:

x

From the diagram above the magnitude of the projection on the plane is:

Calculation of the perpendicular component

|From the above diagram, the scalar magnitude of the perpendicular component
is |A| cos(θ) and its direction is
in the direction of vector B. So if we multiply |A| cos(θ)
by a unit vector along B, which is, B/|B|

So the vector we want is:

A B = |A| cos(θ)
* B/|B|

We can use the vector dot product to calculate this, from this
page we know that:

AB = |A| |B| cos(θ)

Therefore combining these equations gives:

A B = AB * B/|B|2

Alternative using Clifford or Geometric Algebra

For information about Clifford/Geometric Algebra see
here. In Geometric algebra are represented by bivectors.

In this case:

A is a vector

B is a bivector (representing the plane)

Outer product

This is the geometric algebra equivalent of the cross product, but it is not
limited to multiplying vectors by vectors, it increases to grade of operand
as follows:

scalar
vector = vector

vector
vector = bivector

bivector
vector = tri-vector

Inner product

This is the geometric algebra equivalent of the dot product, but it is not
limited to multiplying vectors by vectors, it decreases to grade of operand
as follows:

To replace the dot product the result needs to be a scalar (or a 1×1 matrix which we can get by multiplying by the transpose of B or alternatively just multiply by the scalar factor: (Ax * Bx + Ay * By + Az * Bz)

BA =

Bx

By

Bz

Ax

Ay

Az

perpendicular component

So lets work out the full terms for the perpendicular component:

A B = AB * B / |B|² =

Bx * (Ax * Bx + Ay * By + Az * Bz)
/(Bx² + By² + Bz²)

By * (Ax * Bx + Ay * By + Az * Bz)
/(Bx² + By² + Bz²)

Bz * (Ax * Bx + Ay * By + Az * Bz)
/(Bx² + By² + Bz²)

Assuming B is normalised (Bx² + By² + Bz²=1) and separating out A and B gives:

A B =

Bx²

Bx*By

Bx*Bz

By*Bx

By²

By*Bz

Bz*Bx

Bz*By

Bz²

Ax

Ay

Az

parallel component

We can work out the parallel component in terms of matricies:

A || B = B × (A × B) / |B|²

The cross product anticommutes, that is reversing the order of the operands changes the sign, so we get:

A || B = - B × (B × A) / |B|²

Convert ing this into a matrix equation using the skew symmetric matrix as described above gives:

Where I can, I have put links to Amazon for books that are relevant to
the subject, click on the appropriate country flag to get more details
of the book or to buy it from them.

New Foundations for Classical Mechanics (Fundamental Theories of Physics). This
is very good on the geometric interpretation of this algebra. It has lots of insights
into the mechanics of solid bodies. I still cant work out if the position, velocity,
etc. of solid bodies can be represented by a 3D multivector or if 4 or 5D multivectors
are required to represent translation and rotation.